$\overline{BC} = 10$ $\overline{AC} = {?}$ A C B ? 10 $ \sin( \angle BAC ) = \frac{10\sqrt{181} }{181}, \cos( \angle BAC ) = \frac{9\sqrt{181} }{181}, \tan( \angle BAC ) = \dfrac{10}{9}$
Explanation: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{10}{\overline{AC}} $ $ \overline{AC}=\frac{10}{\tan( \angle BAC )} = \frac{10}{\dfrac{10}{9}} = 9$